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solmaris [256]
2 years ago
8

Tionna's car expenses include car insurance and fuel. Tionna's parents agree to pay a certain percent of her car expenses. The c

ost of car insurance is the same every month, but the cost of fuel depends on the d, number of miles she drives. The expression (0.10d+60)(1−a) represents how much Tionna must pay toward her monthly car expenses.
Which statement is correct?


The expression 1−a represents the percent of car expenses Tionna's parents pay each month, and 60 represents the cost of fuel.

The expression 1−a represents the percent of car expenses Tionna must pay each month, and 60 represents the cost of fuel.

The expression 1−a represents the percent of car expenses Tionna's parents pay each month, and 60 represents the cost of car insurance.

The expression 1−a represents the percent of car expenses Tionna must pay each month, and 60 represents the cost of car insurance.
Mathematics
1 answer:
viva [34]2 years ago
5 0

Answer:

C

Step-by-step explanation:

"The expression 1−a represents the percent of car expenses Tionna's parents pay each month, and 60 represents the cost of car insurance."

You might be interested in
Equivalent fractions for 15/35
Anit [1.1K]
\frac{15}{35} /5 \\  \frac{15/5}{35/5}  \\  \frac{3}{7}

That's the simplest form of 15/35. You can use 3/7 to get other equivalent fractions by multiplying the numerator and denominator by the same number. Other examples are 9/21, 6/14, and 12/28.
Hope this helps!

5 0
3 years ago
A simulation was conducted using 10 fair six-sided dice, where the faces were numbered 1 through 6. respectively. All 10 dice we
kompoz [17]

Answer:

C) a sample distribution of a sample mean with n = 10  

\mu_{{\overline}{X}} = 3.5

and \sigma_{{\overline}{Y}} = 0.38

Step-by-step explanation:

Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .

Let,

X_{ij} = The number which comes up  on the ith die on the jth trial.

∀ i = 1(1)10 and j = 1(1)20

Then,

E(X_{ij}) = \frac {1 + 2 + 3 + 4 + 5 + 6}{6}

                            = 3.5       ∀ i = 1(1)10 and j = 1(1)20

and,

E(X^{2}_{ij} = \frac {1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}

                                = \frac {1 + 4 + 9 + 16 + 25 + 36}{6}

                                = \frac {91}{6}

                                \simeq 15.166667

so, Var(X_{ij} = (E(X^{2}_{ij} - {(E(X_{ij})}^{2})

                                    \simeq 15.166667 - 3.5^{2}

                                    = 2.91667

   and \sigma_{X_{ij}} = \sqrt {2.91667}[/tex                                            [tex]\simeq 1.7078261036

Now we get that,

 Y_{j} = \frac {\sum_{j = 1}^{20}X_{ij}}{20}

We get that Y_{j}'s are iid RV's ∀ j = 1(1)20

Let, {\overline}{Y} = \frac {\sum_{j = 1}^{20}Y_{j}}{20}

      So, we get that E({\overline}{Y}) = E(Y_{j})

                                                                 = E(X_{ij}  for any i = 1(1)10

                                                                 = 3.5

and,

       \sigma_{({\overline}{Y})} = \frac {\sigma_{Y_{j}}}{\sqrt {20}}                                             = \frac {\sigma_{X_{ij}}}{\sqrt {20}}                                             = \frac {1.7078261036}{\sqrt {20}}                                            [tex]\simeq 0.38

Hence, the option which best describes the distribution being simulated is given by,

C) a sample distribution of a sample mean with n = 10  

\mu_{{\overline}{X}} = 3.5

and \sigma_{{\overline}{Y}} = 0.38

                                   

6 0
3 years ago
Solve for a:<br>9+2a = -3-4a​
lubasha [3.4K]

Answer:

a=-2

Step-by-step explanation:

Let me know if you need the steps tho.

5 0
3 years ago
A 51 foot piece of wire is cut into 3 sections so that the first section is three times as long as the second section and the se
kirill [66]

Answer:

The length of the longest section x = 36 ft

Step-by-step explanation:

Total length of the wire = 51 ft

Let first section of wire = x

Second section of wire = y

Third section of wire = z

According to given data

x = 3 y & y = 4 z

Total length of the wire = x + y + z = 51

y + 3 y + \frac{y}{4} = 51

\frac{17}{4} y = 51

y = 12

x = 3 × 12 = 36

z = \frac{y}{4} = \frac{12}{4} = 3

Therefore the length of the longest section x = 36 ft

8 0
3 years ago
Read 2 more answers
Joey Chestnut is trying to break his own world record for eating the most hotdogs and buns in 10 minutes he needs to eat more th
Anna71 [15]
Set up a proportion.
62 over 9 = x over 1
62/9=6.8
9/9=1
6.8 over 1
rounded would be seven
so 6.8 hotdogs per minute
3 0
3 years ago
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